Answer:
In Newton's 1st Law, a stationary body or that moving with a uniform motion, will only change its state if acted upon by other external forces.This restance to change in position/motion is called inertia. In his second law, when a body is acted upon by a force, the force will be equal to the product of mass and acceleration of the body.i.e. Force= m.a. In the third law, it explains that a body will exert an equal and opposite force when intracting with another body.Here, every action force experiences an equal and opposite reaction force, where the momentum of the bodies will be conserved.
Answer:
(e) 3.2
Explanation:
We are given that vector C and D.
Let R be the magnitude of C+D.
According to question
R=3D
We have to find the ratio of the magnitude of C to that of D.
By using right triangle property
Hence, the ratio of the magnitude of C to that of D=3.2
(e) 3.2
Answer:
- Which of the following does not move as a transverse wave?
<em>B. sound waves</em>
because sound waves are longitudinal waves having compressions and rare factions.
Which of these effects describes the change in pitch we hear
moving motorbike goes past?
<em>D</em><em>.</em><em> </em><em>Doppler</em><em> </em><em>eff</em><em>ect</em>
Doppler effect is the phenomenon where there is apparent change in frequency (pitch) and wavelength of a wave due to relative motion of the sound source.
Which of the following does not make use of total internal reflection.
<em>B</em><em>.</em><em> </em><em>Endos</em><em>cop</em><em>e</em>
Endoscope doesn't use total internal reflection since no refraction takes place.
Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
Time it takes the projectile to hit the ground after being thrown up:
√h/1/2a
√8/(.5)(9.81)
√8/4.905
√1.630988787
= 1.277101714
= 1. 28
hope this helps :)
